Part II, Chapter 4Exploring Some Geometric Themes

Every square that remains at Step \(n\) gives rise to 8 squares remaining at Step \(n+1\) (since removing the centre from each 3×3 group leaves 8), so \(R_{n+1}=8R_n\). Starting from \(R_0=1\): \(R_1=8\times1=8\), \(R_2=8\times8=8^2\), and in general:

Every square remaining at Step \(n\) gives rise to exactly 1 new hole at Step \(n+1\), and all holes already present at Step \(n\) remain unchanged. So \(H_{n+1}=H_n+R_n\). Starting from \(H_0=0\): \(H_1=0+1=1\), \(H_2=1+8=9\), \(H_3=9+64=73\), and in general, summing the geometric series \(1+8+8^2+\cdots+8^{n-1}\):

Let the original equilateral triangle have side length \(s\). By the midpoint theorem, the segment joining the midpoints of any two sides is parallel to the third side and exactly half its length. So each of the 3 "corner" triangles formed has all three sides equal to \(\dfrac{s}{2}\) — and since a triangle with two equal sides that are also parallel to (and half of) the original equal sides must have its third side also equal to \(\dfrac{s}{2}\) (using the hint that corner triangles are isosceles, combined with the midpoint theorem), each corner triangle is itself equilateral with side \(\dfrac{s}{2}\).

The central triangle, bounded by the same three midpoint-segments, also has all sides equal to \(\dfrac{s}{2}\) by the same midpoint theorem, so it too is equilateral with side \(\dfrac{s}{2}\).

An \(n\)-sided prism has: \(n\) rectangular/parallelogram side faces plus the 2 congruent \(n\)-gon end faces, giving faces \(=n+2\). Edges: \(n\) edges on each of the two end polygons (\(2n\) total) plus \(n\) vertical edges connecting corresponding vertices, giving edges \(=3n\). Vertices: \(n\) on each end polygon, giving vertices \(=2n\).

An \(n\)-sided pyramid has: \(n\) triangular side faces plus 1 base face, giving faces \(=n+1\). Edges: \(n\) base edges plus \(n\) edges from the apex to each base vertex, giving edges \(=2n\). Vertices: \(n\) base vertices plus 1 apex, giving vertices \(=n+1\).

Take 4 equilateral triangles of the same sidelength (say 6 cm), and arrange 3 of them around a central one, each sharing one full edge with the centre triangle — this is one valid net (as shown above). The second valid net for a regular tetrahedron is a strip of 4 equilateral triangles in a row, alternating "point up, point down."

Draw a square (the base, say 6 cm sides), then attach an isosceles (or equilateral) triangle to each of its 4 sides, all with the same base length as the square's side and pointing outward — folding all 4 triangles upward brings their apexes together to form the pyramid's peak.

If the sector's centre point \(O\) is placed off-centre relative to the circular arc's true centre (i.e. the sector isn't a "clean" portion of a circle centred at \(O\)), folding it up no longer produces a proper cone with a single sharp, symmetric apex — instead, it forms a lopsided, oblique cone-like surface (an "oblique cone"), where the apex isn't directly above the centre of the base circle.

Draw a row of 3 rectangles (matching the prism's length, and each rectangle's width equal to one side of the triangular cross-section) side by side, then attach a triangle (matching the triangular cross-section) to the top edge of one end rectangle, and another identical triangle to the bottom edge of the same or an adjacent rectangle — folding the 3 rectangles around into a triangular tube and folding the 2 triangles in as the end caps completes the prism.

The key trick is to "unfold" the cuboid into a net: a path along the cuboid's surface corresponds exactly to a path of the same length on the net, and vice versa. So the shortest path on the cuboid corresponds to the straight-line path on some valid unfolding of the net between the ant and the laddu — but only if that straight line actually stays within the net's boundary (doesn't cross outside the unfolded shape, which would mean it doesn't correspond to a real path on the solid).

To confirm a candidate path is truly the shortest, check that it becomes a genuine straight line when the surfaces it crosses are unfolded flat — a straight line is always the shortest path between two points on a flat surface, so if the path straightens out perfectly on some valid unfolding, it must be the overall shortest.

Let \(l\) be a line's actual length and \(p\) the length of its projection. Since the projection construction forms a right angle (\(\angle AEB=90^\circ\), where \(AE=p\) is the projected length and \(AB=l\) is the actual length, as the hypotenuse of the right triangle \(AEB\)), \(p\) is one leg of a right triangle with hypotenuse \(l\) — so \(p\le l\) always, since a leg of a right triangle can never exceed its hypotenuse.

A square's projection ranges from a full square (when the square is exactly parallel to the plane), through various rectangles and parallelograms of different proportions (as the square is tilted), down to a single line segment (when the square is viewed exactly edge-on, perpendicular to the plane).

Since parallel lines always project to parallel lines, a parallelogram's two pairs of parallel sides remain parallel in its projection — so the projection of a parallelogram is always another parallelogram (possibly a "squashed" or differently-angled one, or degenerating to a line segment if viewed edge-on), and can never become a non-parallelogram quadrilateral.

Since a polygon's projection is made up of the projections of each of its sides, and each side projects to a (possibly shortened) line segment, the projection of an \(n\)-sided polygon is generally another \(n\)-sided polygon (not necessarily regular — side lengths and angles can change based on orientation). In special orientations, some sides may project onto the same line (if the polygon is tilted so that two sides become collinear in projection), reducing the apparent number of distinct sides, or the whole polygon could collapse to a line segment if viewed exactly edge-on.

A given projection is never made by a unique object — for example, a projected line segment could come from many differently-tilted lines of different lengths (as long as their tilted projections work out to the same length), and a projected square outline of a cuboid could equally come from many different cuboids of different depths (since depth, being along the viewing direction, doesn't show up in the projection at all).

For a cone (whose side-on projection is a triangle), a square pyramid (or any pyramid) oriented appropriately can produce the exact same triangular outline, since the projection only captures the outer silhouette, not the internal edges or the curved vs. flat nature of the surfaces.

For any single line segment in 3D space, its three projected lengths (front, top, side) are each individually \(\le\) the line's actual length (as established earlier: \(p\le l\), with equality only when the line lies exactly parallel to that particular plane). A line oriented purely along one axis will project to its full length in the two views that "see" that axis edge-on, and shrink to a single point in the view looking straight down that axis.

Cube (faces aligned to the 3 planes): front view = square, top view = square, side view = square.

Cuboid (faces aligned to the 3 planes): front view, top view, and side view are each rectangles, generally of 3 different sizes (matching the cuboid's 3 pairs of distinct-sized faces).

Parallelepiped: similar to a cuboid, but the front/top/side views may appear as parallelograms rather than rectangles, depending on how it's tilted.

Cylinder (axis vertical): front view = rectangle, side view = rectangle (same as front view), top view = circle.

Cone (axis vertical, apex up): front view = triangle, side view = triangle (same as front view), top view = circle.

Prism (e.g. triangular prism, lying on a rectangular face, axis horizontal): front view = rectangle, top view = rectangle, side view = the polygon shape of its cross-section (e.g. a triangle).

Pyramid (base horizontal, apex up): front view = triangle, side view = triangle, top view = the polygon shape of its base (e.g. a square, for a square pyramid).

General approach: building a 3D "letter" out of cubes means choosing, for each of the three view directions, which grid positions should be "filled" (visible from that direction) — a cube must be present at every 3D position that is needed to satisfy all three 2D silhouette requirements simultaneously, and no unnecessary cubes should be added elsewhere. Start with the front view's letter shape as a 2D template extruded some fixed depth, then check whether that satisfies the required top view; if not, selectively remove or add cubes at specific depths (without breaking the front view's silhouette) until the top view (and then the side view) also match.

(iv) Other simple block-letter shapes achievable this way include L, T, I, and plus-sign-like shapes, since these have silhouettes that are easy to satisfy simultaneously from multiple viewing directions with a modest number of cubes.

A cube's projection can be: a square (when a face is exactly parallel to the plane); a rectangle (when tilted about an axis parallel to one pair of edges, but not enough to introduce a third visible dimension); a general parallelogram (for other tilts); and a regular hexagon (in the special "isometric" orientation, balanced on one corner vertex, where all edges project to equal lengths).

When a cube is balanced perfectly on one vertex, the 3 edges meeting at the opposite (topmost) vertex are arranged with perfect 3-fold rotational symmetry around the vertical axis through the two opposite corners — by this symmetry, all 3 of those edges (and, by extension, all 12 edges of the cube) must project to exactly the same length, since no direction is treated differently from any other by the symmetric arrangement.

The shadow's shape closely resembles the projection's shape, though the shadow may be scaled up or slightly distorted depending on how close the torch is and how the object is held. As the torch is moved farther away (while staying perpendicular to the wall), the shadow shrinks toward, and becomes indistinguishable from, the true projection — exactly like how sunlight (an effectively infinitely distant, perpendicular light source) casts shadows that are true projections. This gives a simple physical way to verify, for example, that a parallelogram's shadow (and hence its projection) always remains a parallelogram, whatever its orientation.

Yes — the 5 flat Tetris shapes (I, O/square, L, S, and T tetrominoes) all keep the 4 cubes in a single flat layer, but with actual 3D cubes, additional "3D tetromino" (tetracube) combinations become possible by stacking cubes out of that flat plane. For example: an "L-shaped" arrangement where one cube sits on top of another (rather than beside it) in an otherwise flat row of 3, or a "zig-zag staircase" going both sideways and upward, or a 2×2×1 arrangement bent so that one pair of cubes sits at a different height than the other pair.

Yes — this is an "impossible path" illusion, in the style of an Escher never-ending staircase. The arrows suggest a continuous path that keeps going consistently "uphill" (or "downhill") in the picture, when traced all the way around — but a real, physically realisable stack of cubes cannot have a path that keeps rising in height forever and still returns to its starting point at the same height. Only a genuinely realisable portion of the depicted path (following the 3 real, physical primary directions on an actual isometric grid) can be drawn without contradiction; the full loop as pictured relies on the same kind of visual trick used in impossible-staircase artwork.

(i) No — this is the classic "Penrose triangle" illusion. No single real, rigid arrangement of cubes can be built that looks like this from every angle; it only appears to form a closed, continuous triangular loop of cubes when viewed from one very specific viewpoint. Its front, top, and side profiles would each individually look like a perfectly ordinary, simple L-shaped or straight bar of cubes — none of the three individual 2D views look "impossible" on their own; the impossibility only emerges from how our brain wrongly assumes all three connecting corners are simultaneously at consistent depths.

(ii) Drawing this on an isometric grid involves drawing 3 separate L-shaped (or straight) bars of cubes, each oriented along a different one of the 3 principal isometric directions, positioned so their ends appear to overlap and connect at the picture's corners — even though, in true 3D depth, they don't actually meet.

(iii) The illusion works because our visual system automatically (and here, incorrectly) interprets any two line segments that meet at a shared point in a 2D drawing as being physically connected at the same point in 3D space — but the isometric grid allows a single 2D picture to be consistent with the local shape at each corner separately, without any global arrangement of cubes in true 3D actually satisfying all three corners simultaneously.